Let $f(x)=\dfrac{x+1}{xe^{x}+e^{x}}$ when $x\neq -1$. $f$ is continuous for all real numbers. Find $f(-1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $e$ (Choice B) B $1$ (Choice C) C $-2e$ (Choice D) D $2$
$\dfrac{x+1}{xe^{x}+e^{x}}$ is continuous for all real numbers other than $x=-1$ which means $f$ is continuous for all real numbers other than $x=-1$. In order for $f$ to also be continuous at $x=-1$, the following equality must hold: $\lim_{x\to -1}f(x)=f(-1)$ We will obtain the above equality by letting $f(-1)=\lim_{x\to -1}f(x)$. So let's find $\lim_{x\to -1}f(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -1}f(x) \\\\ &=\lim_{x\to -1}\dfrac{x+1}{xe^{x}+e^{x}} \gray{\text{This is the rule for }x\neq -1} \\\\ &=\lim_{x\to -1}\dfrac{\cancel{(x+1)}}{\cancel{(x+1)}(e^{x})} \gray{\text{Factor}} \\\\ &=\lim_{x\to -1}\dfrac{1}{(e^{x})} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq -1) \\\\ &=\dfrac{1}{e^{-1}} \gray{\text{Direct substitution}} \\\\ &=e \end{aligned}$ We obtained that if we set $f(-1)=e$, then $\lim_{x\to -1}f(x)=f(-1)$, which makes $f$ continuous at $x=-1$. Since we already saw that $f$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $f(-1)=e$.